I expressed my desire for a book with the title "(Counter)examples in Algebraic Topology".
My reason for doing so was that while the abstract formalism of algebraic topology is very well-explained in many textbooks and while most graduate students are fond of the general machinery, the study of examples is somehow neglected. I am looking for examples that explain why certain hypotheses are necessary for theorems to hold. The books by Hatcher and Bredon contain some interesting stuff in this direction, and there is Neil Strickland's bestiary, which is mainly focused on positive knowledge.

To convey an idea of what I am after, here are a few examples from my private ''counterexamples in algebraic topology'' list. Some are surprising, some less so.

The abelianization of $SL_2 (Z)$ is $Z/12$, the map $BSL_2(Z) \to BZ/12$ is a homology equivalence to a simple space. But it is not a Quillen plus construction, since the the homotopy fibre is $BF_2$ (free group on $2$ generators), hence not acyclic. See The free group $F_2$ has index 12 in SL(2,$\mathbb{Z}$) .

Self-maps of simply-connected spaces $X$ which are the identity on homotopy, but not on homology (let $X=K(Z;2) \times K(Z;4)$, $u:K(Z;2) \to K(Z;4)$ be the cup square,
and $f:X\to X$ is given by $f(x,y):= (x,y + u(x))$, using that EM-spaces are abelian groups). There are also self-maps of finite simply connected complexes that are the identity on homology, but not on homotopy, see Diarmuid Crowleys answer to Cohomology of fibrations over the circle: how to compute the ring structure?

The stabilization map $B \Sigma_{\infty} \to B \Sigma_{\infty}$ induces a bijection on free homotopy classes $[X, B \Sigma_{\infty}]$ for each finite CW space $X$. However, it is not a homotopy equivalence (not a $\pi_1$-isomorphism).

The fibration $S^1 \to B \mathbb{Q} \to B \mathbb{Q}/\mathbb{Z}$ is classified by a map $f:B \mathbb{Q}/\mathbb{Z} \to CP^{\infty}$, which can be assumed to be a fibration with fibre $B \mathbb{Q}$. Now let $X_n$ be the preimage of the n-skeleton of $CP^{\infty}$. Using the Leray-Serre spectral sequence, we can compute the integral
homology of $X_n$ and, by the universal coefficient theorem, the homology of field coefficients. It turns out that this is finitely generated for any field,
and so we can define the Euler characteristic in dependence of the field. It is not independent of the field in this case (the reason is of course that the integral homology of $X_n$ is not finitely generated).

The compact Lie groups $U(n)$ and $S^1 \times SU(n)$ are diffeomorphic, their classifying spaces have isomorphic cohomology rings and homotopy groups, but the classifying spaces are not homotopy equivalent (look at Steenrod operations).

Question: Which examples of spaces and maps of a similar flavour do you know and want to share with the other MO users?

To focus this question, I suggest to stay in the realm of algebraic topology proper. In other words:

The properties in question should be homotopy invariant properties of spaces/maps. This includes of course fibre bundles, viewed as maps to certain classifying spaces.

Let us talk about spaces of the homotopy type of CW complexes, to avoid that a certain property fails for point-set topological reasons.

This excludes the kind of examples from the famous book "Counterexamples in Topology".

The examples should not be "counterexamples in group theory" in disguise. Any ugly example of a discrete group $G$ gives an equally ugly example of a space $BG$.
Same applies to rings via Eilenberg-Mac Lane spectra.

I prefer examples from unstable homotopy theory.

To get started, here are some questions whose answer I do not know:

Construct two simply-connected CW complexes $X$ and $Y$ such that $H^* (X;F) \cong H^* (Y;F)$ for any field, as rings and modules over the Steenrod algebras, but which are not homotopy equivalent. EDIT: Appropriate Moore spaces do the job, see Eric Wofsey's answer.

Let $f: X \to Y$ be a map of CW-complexes. Assume that $[T,X] \to [T;Y]$ is a bijection for each finite CW complex $T$ ($[T,X]$ denotes free homotopy classes). What assumptions are sufficient to conclude that $f$ is a weak homotopy equivalence? EDIT: the answer has been given by Tyler Lawson, see below.

Do there exist spaces $X,Y,Z$ and a homotopy equivalence $X \times Y \to X \times Z$, without $Y$ and $Z$ being homotopy equivalent? Can I require these spaces to be finite CWs? EDIT: without the finiteness assumptions, this question was ridiculously simple.

Do you know examples of fibrations $F \to E \to X$, such that the integral homology of all three spaces is finitely generated (so that the Euler numbers are defined) and such that the Euler number is not multiplicative, i.e. $\chi(E) \neq \chi(F) \chi(X)$? Remark: is $X$ is assumed to be simply-connected, then the Euler number is multiplicative (absolutely standard). Likewise, if $X$ is a finite CW complex and $F$ is of finite homological type (less standard, but a not so hard exercise). So any counterexample would have to be of infinite type. The above fibration $BSL_2 (Z) \to BZ/12$ is a counterexample away from the primes $2,3$, but do you know one that does the job in all characteristics?. Of course, the ordinary Euler number is the wrong concept here.

I am looking forward for your answers.

EDIT: so far, I have gotten great answers, but mostly for the specific questions I asked. My intention was to create a larger list of counterexamples. So, feel free to mention your favorite strange spaces and maps.

For #1, what about a suspension of a torus versus $S^2 \vee S^2 \vee S^3$? For #3, it's pretty easy to find examples where $X$ is not a finite CW complex (say, an infinite product of $Y \times Z$).
–
John PalmieriFeb 14 '11 at 0:24

5

This should probably be community wiki.
–
Andy PutmanFeb 14 '11 at 1:09

2

For #1, what about a $\mathbf{Z}/4$ Moore space versus a $\mathbf{Z}/8$ Moore space?
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Sam IsaacsonFeb 14 '11 at 1:52

2

For #3: $X$ could be empty. Also, slightly less silly remark, without the finiteness assumption there is the Eilenberg swindle: Let $X$ be the product of infinitely many copies of $Y$ and let $Z$ be a point.
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Tom GoodwillieFeb 14 '11 at 3:46

1

@John: that could be right, Hatcher is using maps $S^{n+m} \to S^n$ which are in the image of the J-homomorphism. If choosen correctlx, these cannot be detected by Steenrod operations.
–
Johannes EbertFeb 14 '11 at 10:15

6 Answers
6

Problem #3 is about noncancellation phenomena. Peter Hilton
(often with G. Mislin and/or J. Roitberg) wrote about ten papers on the subject. There are plenty of examples, generally related to localization.

For #1, an example is given by two Moore spaces $M({\mathbb Z}/p^2,k)$ and $M({\mathbb Z}/p^3,k)$; the only cohomology is in degree $k$ and $k+1$ in characteristic $p$, and the ring and Steenrod module structures are trivial. This works with $p^2$ and $p^3$ replaced by any two powers of $p$ that are not $p$ itself (since for $p$ you get a Bockstein in the cohomology).

The Bockstein is nonzero on $M(Z/p,k)$; that is why you insist on powers larger than one, right?
–
Johannes EbertFeb 14 '11 at 11:40

1

Yes; these spaces can be distinguished by higher cohomology operations. You can produce stable counterexamples to #1 quite generally by looking at the cofibers of maps $S^n \to $S^0$ with Adams filtration greater than 1.
–
Sam IsaacsonFeb 14 '11 at 18:28

This is a great question. Here are two of my favorite counterexamples:

Rector proved in 1971 that there are
uncountably many complexes $X$
(distinct in the homotopy category)
such that $\Omega X \simeq S^3$.

It's possible to construct "ghost" maps $f:X\to Y$ that are zero on $\pi_\ast$, but nonetheless essential (e.g., maps of Eilenberg-Mac Lane spaces representing cohomology operations). Even wilder, there are "phantom" maps $f:X\to Y$ so that if $K$ is any finite complex and $i:K\to X$ a map, $fi \simeq \ast$, but $f \not\simeq \ast$. Dan Christensen has written extensively about phantom maps in the stable homotopy category; I think the first example of a phantom map is due to Adams and Walker.

For (2), a sufficient condition is that the fundamental group $\pi_1(Y)$ be finitely generated. We prove this by showing that the map $X \to Y$ is an isomorphism on homotopy groups.

First, we recall that the set of unbased homotopy classes of maps $[S^n;Z]$ is always isomorphic to the quotient $\pi_n(Z)/\pi_1(Z)$ of the set of based maps by the action of the fundamental group. The identity element $e \in \pi_n(Z)$ is always a singleton orbit. Therefore, taking $T = S^n$ we find that the bijection $[S^n;X] \to [S^n;Y]$ implies that the kernel of the map $\pi_n(X) \to \pi_n(Y)$ is trivial. Therefore, the map is automatically injective on homotopy groups.

Even further, this allows us to reduce to showing that the map $\pi_1(X) \to \pi_1(Y)$ is surjective, as follows. For any $y \in \pi_n(Y)$, the conjugacy class $[y]$ is in the image, so there exists an element $x \in \pi_n(X)$ whose image is ${}^g y$ for some $g \in \pi_1(Y)$. Surjectivity on $\pi_1$ would imply that we can lift $g^{-1}$ to a class in $h \in \pi_1(X)$, and then the image of ${}^h x$ is $y$.

In order to prove surjectivity, we recall a similar result about maps from wedges of circles. A based map $\vee^k S^1 \to Z$ determines an ordered k-tuple of elements of $\pi_1 Z$, and the set of unbased maps $[\vee^k S^1;Z]$ is the set of simultaneous conjugacy classes of such k-tuples.

If $\pi_1(Y)$ is finitely generated, construct a based map $\vee^k S^1 \to Y$ corresponding to a k-tuple $(y_1, \ldots, y_k)$ of elements generating the group. Since the map $[\vee^k S^1; X] \to [\vee^k S^1; Y]$ is a bijection, the simultaneous conjugacy class is in the image; i.e. there exists an element $g \in \pi_1(Y)$ such that the k-tuple $(gy_1 g^{-1}, gy_2 g^{-1}, \ldots, gy_k g^{-1})$ is in the image. However, this is another set of generators for $\pi_1(Y)$, and so the map $\pi_1(X) \to \pi_1(Y)$ is surjective, as desired.

Great! So, by the first two paragraphs, $\pi_1$-surjectivity is sufficient. This is also guaranteed if $\pi_1 (Y)$ is abelian.
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Johannes EbertFeb 14 '11 at 17:59

Yes, that's right. So counterexamples must fall into the "infinitely generated nonabelian fundamental group" category, which is a little more pathological than "non-simply-connected".
–
Tyler LawsonFeb 14 '11 at 20:22

Also, it appears that I was lazy and assumed that X and Y are path-connected throughout, which is a case that you can reduce to using $\pi_0$.
–
Tyler LawsonFeb 14 '11 at 20:22

What follows is merely a reference to the excellent answer and comment by Karol Szumiło in this mathoverflow question asked by Mike Shulman. There, Karol provides arguments and bibliographic sources which prove the failure of Brown representability in the homotopy category of unpointed spaces, and in the homotopy category of not necessarily connected pointed spaces. Please upvote Karol's comment and answer, together with the question. For convenience, I review below the main points of that discussion.$\newcommand{\Ho}{\mathrm{Ho}}
\newcommand{\op}[1]{#1^{\mathrm{op}}}
\newcommand{\Set}{\mathrm{Set}}$

In the article "Splitting homotopy idempotents II", Peter Freyd and Alex Heller construct a very special homotopy idempotent $Bf:BF\to BF$, i.e. $Bf$ is an idempotent in $\Ho$, the homotopy category of spaces with the homotopy type of CW-complexes. Here, $F$ denotes Thompson's group, and $BF$ is its classifying space. Importantly, the homotopy idempotent $Bf:BF\to BF$ does not split, i.e. does not admit a retract in the homotopy category. This example was also constructed independently by Dydak in his 1977 article "A simple proof that pointed connected FANR-spaces are regular fundamental retracts of ANR's".

The idempotent $Bf:BF\to BF$ provides a retract $R$ of the representable functor $[-,BF]:\op{\Ho}\to\Set$, since idempotents do split in the category $\Set$ of sets. Then $R$ is necessarily half-exact, i.e. preserves small products and weak pullbacks; these are the conditions in Brown's representability theorem. However, $R$ is not representable since a representing object for $R$ would be a retract for $Bf:BF\to BF$ in $\Ho$. We can also apply the same argument to the pointed map $(Bf)_+ : (BF)_+ \to (BF)_+$ obtained by adding disjoint basepoints, giving a half-exact, non-representable functor $\op{(\Ho_\ast)}\to\Set$ on the pointed homotopy category.

More examples of the failure of Brown representability are described by Alex Heller in the article "On the representability of homotopy functors", which provides some more fascinating insights into the phenomenon of Brown representability. In particular, that article gives a functor $N:\op{\Ho}\to\Set$ which is half-exact, but is not even a retract of any representable functor. The functor $N$ is defined for each space $X$ by
$$ N(X)=\prod_{[x]\in\pi_0(X)} S\bigl(\pi_1(X,x)\bigr) $$
where $S(G)$ is the set of normal subgroups of $G$, for each group $G$. Observe that the choice of $x\in X$ representing $[x]\in\pi_0 X$ is inconsequential. For a homotopy class of maps $[f]:X\to Y$ in $\Ho$, the function $N([f])$ is given by taking preimages of normal subgroups by $f_\ast:\pi_1(X,x)\to\pi_1(Y,f(x))$. Then $N$ is not a retract of a representable functor because normal subgroups of $\pi_1(X,x)$ can have arbitrarily large index if one varies the space $X$.